Problem: $\dfrac{ -3j + 3k }{ -10 } = \dfrac{ 7j + 5l }{ 2 }$ Solve for $j$.
Answer: Multiply both sides by the left denominator. $\dfrac{ -3j + 3k }{ -{10} } = \dfrac{ 7j + 5l }{ 2 }$ $-{10} \cdot \dfrac{ -3j + 3k }{ -{10} } = -{10} \cdot \dfrac{ 7j + 5l }{ 2 }$ $-3j + 3k = -{10} \cdot \dfrac { 7j + 5l }{ 2 }$ Reduce the right side. $-3j + 3k = -{10} \cdot \dfrac{ 7j + 5l }{ {2} }$ $-3j + 3k = -{5} \cdot \left( 7j + 5l \right)$ Distribute the right side $-3j + 3k = -{5} \cdot \left( {7j} + {5l} \right)$ $-3j + 3k = -{35}j - {25}l$ Combine $j$ terms on the left. $-{3j} + 3k = -{35j} - 25l$ ${32j} + 3k = -25l$ Move the $k$ term to the right. $32j + {3k} = -25l$ $32j = -25l - {3k}$ Isolate $j$ by dividing both sides by its coefficient. ${32}j = -25l - 3k$ $j = \dfrac{ -25l - 3k }{ {32} }$